hdu 4362 Dragon Ball(DP+优化，4级)

最新推荐文章于 2019-10-30 14:44:45 发布

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最新推荐文章于 2019-10-30 14:44:45 发布

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原文链接：http://www.cnblogs.com/nealgavin/p/3204864.html

本文介绍了一种解决寻找龙珠最小能量消耗问题的算法。通过动态规划和二分查找的方法，在限定的时间和能量下，计算出获取所有时间段内出现的龙珠所需的最低能量消耗。

C - Dragon Ball

Time Limit:1500MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4362

Description

Sean has got a Treasure map which shows when and where the dragon balls will appear. some dragon balls will appear in a line at the same time for each period.Since the time you got one of them,the other dragon ball will disappear so he can only and must get one Dragon ball in each period.Digging out one ball he will lose some energy.Sean will lose |x-y| energy when he move from x to y.Suppose Sean has enough time to get any drogan ball he want in each period.We want to know the minimum energy sean will lose to get all period’s dragon ball.

Input

In the first line a number T indicate the number of test cases.Then for each case the first line contain 3 numbers m,n,x(1<=m<=50,1<=n<=1000),indicate m period Dragon ball will appear,n dragon balls for every period, x is the initial location of sean.Then two m*n matrix. For the first matrix,the number in I row and J column indicate the location of J-th Dragon ball in I th period.For the second matrix the number in I row and J column indicate the energy sean will lose for J-th Dragon ball in I-th period.

Output

For each case print a number means the minimum energy sean will lose.

Sample Input

      1 3 2 5 2 3 4 1 1 3 1 1 1 3 4 2
     

Sample Output

8

思路：dp[i][j]=min{dp[i-1][k]+abs(arr[i][j].x-arr[i-1][k].x)+arr[i][j].w}

二分法

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
const int mm=1100;
const int oo=1e9;
int dp[55][mm];
int cas;
int ll[mm],rr[mm];
int fabs(int x)
{
  if(x<0)return -x;
  return x;
}
class node
{
  public:int wei,w;
  bool operator <(const node&x)const
  {
    if(wei==x.wei)return w<x.w;
    return wei<x.wei;
  }
}f[55][mm];
int Q[mm],pos;
int look(int x,int l,int r,int num)
{ int mid;
  while(l<=r)
  {
    mid=(l+r)/2;
    if(f[x][mid].wei<num)l=mid+1;
    else r=mid-1;
  }
  return r;
}
int main()
{ int n,m,x;
  while(~scanf("%d",&cas))
  {
    while(cas--)
    {
      scanf("%d%d%d",&m,&n,&x);
      for(int i=0;i<m;++i)
        for(int j=0;j<n;++j)
        scanf("%d",&f[i][j].wei);
      for(int i=0;i<m;++i)
        for(int j=0;j<n;j++)
        scanf("%d",&f[i][j].w);
      for(int i=0;i<m;++i)
        sort(f[i],f[i]+n);
      for(int i=0;i<n;++i)
        dp[0][i]=f[0][i].w+fabs(f[0][i].wei-x);
      for(int i=1;i<m;++i)
      { int _min=oo;
        for(int j=0;j<n;++j)
        { _min=min(_min,dp[i-1][j]-f[i-1][j].wei);
          ll[j]=_min;
        }
        _min=oo;
        for(int j=n-1;j>=0;--j)
        {
          _min=min(_min,dp[i-1][j]+f[i-1][j].wei);
          rr[j]=_min;
        }
        for(int j=0;j<n;++j)
        {
          dp[i][j]=oo;
          if(f[i][j].wei<=f[i-1][0].wei)
            dp[i][j]=min(dp[i][j],rr[0]-f[i][j].wei+f[i][j].w);///左边大
          else if(f[i][j].wei>=f[i-1][n-1].wei)
            dp[i][j]=min(dp[i][j],ll[n-1]+f[i][j].wei+f[i][j].w);
          else
          {
            int xx=look(i-1,0,n-1,f[i][j].wei);///比其小的第一个
            dp[i][j]=min(dp[i][j],min(ll[xx]+f[i][j].wei+f[i][j].w,rr[xx+1]-f[i][j].wei+f[i][j].w));
          }
        }
      }
      int ans=oo;
      for(int i=0;i<n;++i)
        ans=min(dp[m-1][i],ans);
        printf("%d\n",ans);
    }
  }
  return 0;
}

优先队列

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
const int mm=1100;
const int oo=1e9;
int dp[55][mm];
int cas;
int fabs(int x)
{
  if(x<0)return -x;
  return x;
}
class node
{
  public:int wei,w;
  bool operator <(const node&x)const
  {
    if(wei==x.wei)return w<x.w;
    return wei<x.wei;
  }
}f[55][mm];
int Q[mm],pos;
int main()
{ int n,m,x;
  while(~scanf("%d",&cas))
  {
    while(cas--)
    {
      scanf("%d%d%d",&m,&n,&x);
      for(int i=0;i<m;++i)
        for(int j=0;j<n;++j)
        scanf("%d",&f[i][j].wei);
      for(int i=0;i<m;++i)
        for(int j=0;j<n;j++)
        scanf("%d",&f[i][j].w);
        for(int i=0;i<=m;++i)
          for(int j=0;j<=n;++j)
          dp[i][j]=oo;
      for(int i=0;i<m;++i)
        sort(f[i],f[i]+n);
      for(int i=0;i<n;++i)
        dp[0][i]=f[0][i].w+fabs(f[0][i].wei-x);
      for(int i=1;i<m;++i)
      {  pos=0;
        Q[pos++]=oo;int k=0;
        for(int j=0;j<n;++j)
        {
         while(k<n)
         {
           if(f[i][j].wei<f[i-1][k].wei)break; ///当前位置比上一层大
           while(pos)
           {
             if(Q[pos-1]>dp[i-1][k]-f[i-1][k].wei)
              --pos;
             else break;
           } Q[pos++]=dp[i-1][k]-f[i-1][k].wei;
             ++k;
         }
         dp[i][j]=Q[0]+f[i][j].w+f[i][j].wei;
        }
         pos=0;
        Q[pos++]=oo;k=n-1;
        /*dp[i][j]=min(dp[i-1][j]+f[i-1][k].wei-f[i][j].wei+f[i][j].w)*/
        for(int j=n-1;j>=0;--j)
        {
         while(k>=0)
         {
           if(f[i][j].wei>f[i-1][k].wei)break;///当前位置要比上一层小
           while(pos)
           {
             if(Q[pos-1]>dp[i-1][k]+f[i-1][k].wei)///找到比当前小的为值，因为比它大的肯不是最优
              --pos;
             else break;
           } Q[pos++]=dp[i-1][k]+f[i-1][k].wei;
             --k;
         }
         dp[i][j]=min(dp[i][j],Q[0]-f[i][j].wei+f[i][j].w);
        }
      }
      //puts("++++");
     // for(int i=0;i<m;++i)
       // for(int j=0;j<n;++j)
        //printf("%d%c",dp[i][j],j==n-1?'\n':' ');
      int ans=oo;
      for(int i=0;i<n;++i)
        ans=min(dp[m-1][i],ans);
        printf("%d\n",ans);
    }
  }
  return 0;
}

转载于:https://www.cnblogs.com/nealgavin/p/3204864.html